Problem: What is the extraneous solution to these equations? $\dfrac{x^2 - 3x}{x + 10} = \dfrac{-10x + 30}{x + 10}$
Multiply both sides by $x + 10$ $ \dfrac{x^2 - 3x}{x + 10} (x + 10) = \dfrac{-10x + 30}{x + 10} (x + 10)$ $ x^2 - 3x = -10x + 30$ Subtract $-10x + 30$ from both sides: $ x^2 - 3x - (-10x + 30) = -10x + 30 - (-10x + 30)$ $ x^2 - 3x + 10x - 30 = 0$ $ x^2 + 7x - 30 = 0$ Factor the expression: $ (x + 10)(x - 3) = 0$ Therefore $x = -10$ or $x = 3$ At $x = -10$ , the denominator of the original expression is 0. Since the expression is undefined at $x = -10$, it is an extraneous solution.